Student Attendance Record I

Difficulty: Easy | Acceptance: 50.20% | Paid: No Topics: String

You are given a string s representing an attendance record for a student where each character signifies whether the student was absent, late, or present on that day. The record only contains the following three characters:

‘A’: Absent. ‘L’: Late. ‘P’: Present.

The student is eligible for an attendance award if they meet both of the following criteria:

The student was absent (‘A’) for strictly fewer than 2 days total. The student was never late (‘L’) for 3 or more consecutive days. Return true if the student is eligible for an attendance award, or false otherwise.

• Examples
• Constraints
• Single Pass with Counters
• String Methods Approach
• Regular Expression Approach

Examples

Input: s = "PPALLP"
Output: true
Explanation: The student has fewer than 2 absences and was never late 3 or more consecutive days.

Input: s = "PPALLL"
Output: false
Explanation: The student was late 3 consecutive days in the last 3 days.

Constraints

1 <= s.length <= 1000
s[i] is either 'A', 'L', or 'P'.

Single Pass with Counters

Intuition Iterate through the string once, tracking the total count of absences and the current streak of consecutive late days.

Steps

• Initialize counters for absences and consecutive late days
• For each character, update the appropriate counter
• Reset consecutive late counter when we see ‘A’ or ‘P’
• Return false early if either condition is violated
• Return true if we complete the iteration without violations

class Solution:
    def checkRecord(self, s: str) -> bool:
        count_A = 0
        consecutive_L = 0
        
        for char in s:
            if char == 'A':
                count_A += 1
                consecutive_L = 0
                if count_A >= 2:
                    return False
            elif char == 'L':
                consecutive_L += 1
                if consecutive_L >= 3:
                    return False
            else:  # 'P'
                consecutive_L = 0
        
        return True

Complexity

• Time: O(n) where n is the length of the string
• Space: O(1) only using constant extra space
• Notes: Early termination possible when conditions are violated

String Methods Approach

Intuition Use built-in string methods to directly check the two conditions: count of ‘A’ and presence of “LLL” substring.

Steps

• Count occurrences of ‘A’ using count method
• Check if “LLL” exists in the string using contains or includes
• Return true only if count of ‘A’ is less than 2 AND “LLL” is not present

class Solution:
    def checkRecord(self, s: str) -> bool:
        return s.count('A') < 2 and 'LLL' not in s

Complexity

• Time: O(n) for counting and substring search
• Space: O(1) constant extra space
• Notes: More readable but may traverse string multiple times internally

Regular Expression Approach

Intuition Use regex pattern matching to check for invalid patterns: two or more ‘A’s or three consecutive ‘L’s.

Steps

• Create regex pattern to match two ‘A’s with any characters in between
• Create regex pattern to match three consecutive ‘L’s
• Return true if neither pattern is found in the string

import re

class Solution:
    def checkRecord(self, s: str) -> bool:
        return not (re.search(r'A.*A', s) or re.search(r'LLL', s))

Complexity

• Time: O(n) regex matching complexity
• Space: O(1) constant extra space for patterns
• Notes: Elegant but regex overhead may be slower for simple patterns